Question

An analytical chemist is titrating 114.8mL of a 0.3100M solution of nitrous acid HNO2 with a 1.000M solution of KOH . The pKa of nitrous acid is 3.35 . Calculate the pH of the acid solution after the chemist has added 12.69mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places.

Answer 1

Ans :- pH = 3.1

Explanation :-

Number of moles of HNO₂ = Molarity x Volume = 0.3100 M x 0.1148 L = 0.03559 mol

Number of moles of KOH = 1.000 M x 0.01269 L = 0.01269 mol

Reaction between KOH and HNO₂ is :

.........................HNO₂ (aq)............+............OH^- (aq) <---------------> NO₂^- (aq)................+..............H₂O (l)

Initial (I).............0.03559 mol.......................0.01269 mol.........................0.0 mol.......................................

Change (C).......-0.01269............................-0.01269............................+0.01269.......................................

Equilibrium (E)...0.0229 mol...........................0.0 mol...............................0.01269 mol

Because, resulting solution of HNO₂ acid and its NO₂^- conjugate base forms an acidic buffer solution and we knows that the pH of the acidic buffer solution can be calculated by using Henderson-Hasselbalch equation i.e.

pH = pK_a + log [Conjugate base] / [acid]

pH = 3.35 + log [NO₂^-] / [HNO₂]

pH = 3.35 + log 0.01269 / 0.0229

pH = 3.35 + log 0.5541

pH = 3.35 - 0.2564

pH = 3.09436

pH = 3.1